I'm having issues proving that the multiplicative identity is unique on the integers. Heres what I have so far,
EDIT:
Suppose $\exists \ \theta_{1},\theta_{2} \ such \ that \ \theta_{1} \neq \theta_{2}$
$\theta_{1} = [(x+1,x)]$
$\theta_{2} = [(y+1,y)]$
$\theta_1 \otimes \theta_2 = [(m+1,m)] \otimes [(k+1,k)]$
$= [((m+1)(k+1) + (m)(k),(m)(k+1)+(m+1)(k))]$
$= [(mk + m + k + 1 + mk, mk + m + mk + k)]$
Since,
$mk + m + k + 1 + mk + m = mk + m + mk + k + m + 1$
We can say that
$(mk + m + k + 1 + mk, mk + m + mk + k)$ ~ $(m + 1,m)$
Thus $\theta_1 = \theta_1 \otimes \theta_2$
Similar argument for $\theta_2$
How do I go about showing that $\theta_1 = \theta_2$?
HINT $$\theta_1 = \theta_1 \otimes \theta_2 = \theta_2$$